Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 73


$\dfrac{\sqrt 3}{2}$

Work Step by Step

The reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps: a) Quadrant- I: $\theta $ b) Quadrant- II: $180^{\circ}-\theta $ c) Quadrant -III: $\theta - 180^o $ d) Quadrant -IV: $360^{\circ}-\theta $ Reference angle of $240^{\circ}$ is equal to $ =240^{\circ}-180^{\circ}=60^{\circ}$ Since, $\sin 60^{\circ}=\dfrac{\sqrt 3}{2}$ So, $\sin (-240)^{\circ}= \dfrac{\sqrt 3}{2}$; Because $\theta $ lies in Quadrant-II.
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